# How do you find the value of the discriminant and determine the nature of the roots 2 x^2 - 3x +1=0?

Mar 13, 2018

Solution: $x = 1 \mathmr{and} x = \frac{1}{2}$

#### Explanation:

$2 {x}^{2} - 3 x + 1 = 0$

Comparing with standard quadratic equation $a {x}^{2} + b x + c = 0$

$a = 2 , b = - 3 , c = 1$ Discriminant, $D = {b}^{2} - 4 a c$ or

$D = 9 - 8 = 1$. If discriminant positive, we get two real solutions,

if it is zero we get just one solution, and if it is negative we get

complex roots. Discriminant is positive here , so it has two

distinct root . Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{3 \pm \sqrt{1}}{4} \therefore x = \frac{3 \pm 1}{4} \therefore x = 1 \mathmr{and} x = \frac{1}{2}$

Solution: $x = 1 \mathmr{and} x = \frac{1}{2}$ [Ans]